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A concrete piece of evidence for incompleteness

On Thursday, the 25th of March 2015, Venanzio Capretta gave a Theory Lunch talk about Goodstein’s theorem. Later, on the 9th of March, Wolfgang Jeltsch talked about ordinal numbers, which are at the base of Goodstein’s proof. Here, I am writing down a small recollection of their arguments.

Given a base , consider the base- writing of the nonnegative integer

where each is an integer between and . The Cantor base- writing of is obtained by iteratively applying the base- writing to the exponents as well, until the only values appearing are integers between and . For example, for and , we have

and also

Given a nonnegative integer , consider the Goodstein sequence defined for by putting , and by constructing from as follows:

Take the Cantor base- representation of .

Convert each into , getting a new number.

If the value obtained at the previous point is positive, then subtract from it.
(This is called the woodworm’s trick.)

Goodstein’s proof relies on the use of ordinal arithmetic. Recall the definition: an ordinal number is an equivalence class of well-ordered sets modulo order isomorphisms, i.e., order-preserving bijections.Observe that such order isomorphism between well-ordered sets, if it exists, is unique: if and are well-ordered sets, and are two distinct order isomorphisms, then either or has a minimum , which cannot correspond to any element of .

An interval in a well-ordered set is a subset of the form .

Fact 1. Given any two well-ordered sets, either they are order-isomorphic, or one of them is order-isomorphic to an initial interval of the other.

In particular, every ordinal is order-isomorphic to the interval .

All ordinal numbers can be obtained via von Neumann’s classification:

The zero ordinal is , which is trivially well-ordered as it has no nonempty subsets.

A successor ordinal is an ordinal of the form , with every object in being smaller than in .
For instance, can be seen as .

A limit ordinal is a nonzero ordinal which is not a successor. Such ordinal must be the least upper bound of the collection of all the ordinals below it.
For instance, the smallest transfinite ordinal is the limit of the collection of the finite ordinals.

Observe that, with this convention, each ordinal is an element of every ordinal strictly greater than itself.

Fact 2. Every set of ordinal numbers is well-ordered with respect to the relation: if and only if .

Operations between ordinal numbers are defined as follows: (up to order isomorphisms)

is a copy of followed by a copy of , with every object in being strictly smaller than any object in .
If and are finite ordinals, then has the intuitive meaning. On the other hand, , as a copy of followed by a copy of is order-isomorphic to : but is strictly larger than , as the latter is an initial interval of the former.

is a stack of copies of , with each object in each layer being strictly smaller than any object of any layer above.
If and are finite ordinals, then has the intuitive meaning. On the other hand, is a stack of copies of , which is order-isomorphic to : but is a stack of copies of , which is order-isomorphic to .

is if , if is the successor of , and the least upper bound of the ordinals of the form with if is a limit ordinal.
If and are finite ordinals, then has the intuitive meaning. On the other hand, is the least upper bound of all the ordinals of the form where is a finite ordinal, which is precisely : but .

Proof of Goodstein’s theorem: To each integer value we associate an ordinal number by replacing each (which, let’s not forget, is the base is written in) with . For example, if , then

and (which, incidentally, equals ) so that

We notice that, in our example, , but : why is it so?, and is it just a case, or is there a rule behind this?

At each step where , consider the writing . Three cases are possible:

.
Then , as , and .

and .
Then for a transfinite ordinal , and .

and .
Then for some , and is a number whose th digit in base is zero: correspondingly, the rightmost term in will be replaced by a smaller ordinal in .

It is then clear that the sequence is strictly decreasing. But the collection of all ordinals not larger than is a well-ordered set, and every nonincreasing sequence in a well-ordered set is ultimately constant: hence, there must be a value such that . But the only way it can be so is when : in turn, the only option for to be zero, is that is zero as well. This proves the theorem.

So why is it that Goodstein’s theorem is not provable in the first order Peano arithmetics? The intuitive reason, is that the exponentiations can be arbitrarily many, which requires having available all the ordinals up to

, times , times:

this, however, is impossible if induction only allows finitely many steps, as it is the case for first-order Peano arithmetics. A full discussion of a counterexample, however, would greatly exceed the scope of this post.